# Integration of Hyperbolic Functions

## Solved Problems

Click or tap a problem to see the solution.

### Example 9

Evaluate the integral $\int {{e^{-x}}\sinh 2xdx}.$

### Example 10

Evaluate the integral $\int {\frac{{dx}}{{\sinh x}}}.$

### Example 11

Find the integral $\int {\frac{{dx}}{{1 + \cosh x}}}.$

### Example 12

Find the integral $\int {\frac{{dx}}{{\sinh x + 2\cosh x}}}.$

### Example 13

Find the integral $\int {\frac{{dx}}{{\sinh x - \cosh x}}}.$

### Example 14

Find the integral $\int {\frac{{dx}}{{3\sinh x - 5\cosh x}}}.$

### Example 15

Evaluate the integral $\int {\sinh x\cosh \left( { - x} \right)dx}.$

### Example 16

Calculate the integral $\int {\sinh 2x\cosh 3xdx}.$

### Example 17

Find the integral $\int {\tanh 2xdx}.$

### Example 18

Find the integral $\int {\coth \frac{x}{3} dx}.$

### Example 19

Evaluate the integral $\int {\sinh x\cos xdx}.$

### Example 9.

Evaluate the integral $\int {{e^{-x}}\sinh 2xdx}.$

Solution.

The hyperbolic sine function is defined as

$\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}.$

Hence the integral is given by

$\int {{e^{ - x}}\sinh 2xdx} = \int {{e^{ - x}}\left( {\frac{{{e^{2x}} - {e^{ - 2x}}}}{2}} \right)dx} = \frac{1}{2}\int {\left( {{e^x} - {e^{ - 3x}}} \right)dx} = \frac{1}{2}\left[ {{e^x} - \frac{{{e^{ - 3x}}}}{{\left( { - 3} \right)}}} \right] + C = \frac{{{e^x}}}{2} + \frac{{{e^{ - 3x}}}}{6} + C.$

### Example 10.

Evaluate the integral $\int {\frac{{dx}}{{\sinh x}}}.$

Solution.

Using the identity

${\cosh ^2}x - {\sinh ^2}x = 1,$

we rewrite the integral as follows:

$I = \int {\frac{{dx}}{{\sinh x}}} = \int {\frac{{\sinh x}}{{{{\sinh }^2}x}}dx} = \int {\frac{{\sinh x}}{{{{\cosh }^2}x - 1}}dx} .$

Make the substitution

$u = \cosh x,\;\; \Rightarrow du = \sinh xdx.$

Hence

$I = \int {\frac{{du}}{{{u^2} - 1}}} = \frac{1}{2}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C = \frac{1}{2}\ln \left| {\frac{{\cosh x - 1}}{{\cosh x + 1}}} \right| + C = \ln \sqrt {\left| {\frac{{\cosh x - 1}}{{\cosh x + 1}}} \right|} + C.$

We can simplify the expression under the square root. Since

$\cosh x - 1 = {\cosh ^2}\frac{x}{2} + {\sinh ^2}\frac{x}{2} - \left( {{{\cosh }^2}\frac{x}{2} - {{\sinh }^2}\frac{x}{2}} \right) = \cancel{{\cosh ^2}\frac{x}{2}} + {\sinh ^2}\frac{x}{2} - \cancel{{\cosh ^2}\frac{x}{2}} + {\sinh ^2}\frac{x}{2} = 2\,{\sinh ^2}\frac{x}{2},$
$\cosh x + 1 = {\cosh ^2}\frac{x}{2} + \cancel{{\sinh ^2}\frac{x}{2}} + {\cosh ^2}\frac{x}{2} - \cancel{{\sinh ^2}\frac{x}{2}} = 2\,{\cosh ^2}\frac{x}{2},$

we obtain

$\frac{{1 - \cosh x}}{{1 + \cosh x}} = \frac{{\cancel{2}{{\sinh }^2}\frac{x}{2}}}{{\cancel{2}{{\cosh }^2}\frac{x}{2}}} = {\tanh ^2}\frac{x}{2}.$

The integral is expressed in the form

$I = \ln \sqrt {\left| {\frac{{\cosh x - 1}}{{\cosh x + 1}}} \right|} + C = \ln \sqrt {\left| {{{\tanh }^2}\frac{x}{2}} \right|} + C = \ln \left| {\tanh \frac{x}{2}} \right| + C.$

### Example 11.

Find the integral $\int {\frac{{dx}}{{1 + \cosh x}}}.$

Solution.

By definition of the hyperbolic cosine,

$\cosh x = {\frac{{{e^x} + {e^{ - x}}}}{2}}.$

Hence, the integral is equal

$\int {\frac{{dx}}{{1 + \cosh x}}} = \int {\frac{{dx}}{{1 + \frac{{{e^x} + {e^{ - x}}}}{2}}}} = \int {\frac{{2dx}}{{2 + {e^x} + {e^{ - x}}}}} = 2\int {\frac{{{e^x}dx}}{{2{e^x} + {e^{2x}} + 1}}} = 2\int {\frac{{d\left( {{e^x} + 1} \right)}}{{{{\left( {{e^x} + 1} \right)}^2}}}} = - \frac{2}{{{e^x} + 1}} + C.$

### Example 12.

Find the integral $\int {\frac{{dx}}{{\sinh x + 2\cosh x}}}.$

Solution.

By definition,

$\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} \;\;\text{and}\;\; \cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}.$

Hence,

$I = \int {\frac{{dx}}{{\sinh x + 2\cosh x}}} = \int {\frac{{dx}}{{\frac{{{e^x} - {e^{ - x}}}}{2} + 2 \cdot \frac{{{e^x} + {e^{ - x}}}}{2}}}} = \int {\frac{{2dx}}{{{e^x} - {e^{ - x}} + 2{e^x} + 2{e^{ - x}}}}} = 2\int {\frac{{dx}}{{3{e^x} + {e^{ - x}}}}} = 2\int {\frac{{{e^x}dx}}{{3{e^{2x}} + 1}}} .$

We make the substitution: $$u = {e^x},$$ $$du = {e^x}dx$$ and calculate the initial integral:

$I = 2\int {\frac{{{e^x}dx}}{{3{e^{2x}} + 1}}} = 2\int {\frac{{du}}{{3{u^2} + 1}}} = \frac{2}{3}\int {\frac{{du}}{{{u^2} + \frac{1}{3}}}} = \frac{2}{3}\int {\frac{{du}}{{{u^2} + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}} = \frac{2}{3} \cdot \frac{{\sqrt 3 }}{1}\arctan \frac{u}{{\frac{1}{{\sqrt 3 }}}} + C = \frac{2}{{\sqrt 3 }}\arctan \left( {\sqrt 3 u} \right) + C = \frac{2}{{\sqrt 3 }}\arctan \left( {\sqrt 3 {e^x}} \right) + C.$

### Example 13.

Find the integral $\int {\frac{{dx}}{{\sinh x - \cosh x}}}.$

Solution.

Since

$\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2},\;\;\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2},$

the integral is expressed in the form

$\int {\frac{{dx}}{{\sinh x - \cosh x}}} = \int {\frac{{dx}}{{\frac{{{e^x} - {e^{ - x}}}}{2} - \frac{{{e^x} + {e^{ - x}}}}{2}}}} = \int {\frac{{2dx}}{{\cancel{e^x} - {e^{ - x}} - \cancel{e^x} - {e^{ - x}}}}} = \int {\frac{{\cancel{2}dx}}{{ - \cancel{2}{e^{ - x}}}}} = - \int {{e^x}dx} = - {e^x} + C.$

### Example 14.

Find the integral $\int {\frac{{dx}}{{3\sinh x - 5\cosh x}}}.$

Solution.

We substitute the definitions of the hyperbolic sine and cosine functions in the integrand:

$\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2},\;\;\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}.$

This yields:

$I = \int {\frac{{dx}}{{3\sinh x - 5\cosh x}}} = \int {\frac{{dx}}{{3 \cdot \frac{{{e^x} - {e^{ - x}}}}{2} - 5 \cdot \frac{{{e^x} + {e^{ - x}}}}{2}}}} = \int {\frac{{2dx}}{{3{e^x} - 3{e^{ - x}} - 5{e^x} - 5{e^{ - x}}}}} = \int {\frac{{2dx}}{{ - 2{e^x} - 8{e^{ - x}}}}} = - \int {\frac{{dx}}{{{e^x} + 4{e^{ - x}}}}} = - \int {\frac{{dx}}{{{e^x} + 4{e^{ - x}}}}} .$

Next, we multiply the numerator and denominator by $${e^x}:$$

$I = - \int {\frac{{dx}}{{{e^x} + 4{e^{ - x}}}}} = - \int {\frac{{{e^x}dx}}{{{e^{2x}} + 4}}} ,$

and make the substitution

$u = {e^x},\;\; \Rightarrow du = {e^x}dx.$

We obtain a basic integral:

$I = - \int {\frac{{{e^x}dx}}{{{e^{2x}} + 4}}} = - \int {\frac{{du}}{{{u^2} + {2^2}}}} = - \frac{1}{2}\arctan \frac{u}{2} + C = - \frac{1}{2}\arctan \frac{{{e^x}}}{2} + C.$

### Example 15.

Evaluate the integral $\int {\sinh x\cosh \left( { - x} \right)dx}.$

Solution.

The hyperbolic cosine function is even:

$\cosh \left( { - x} \right) = \cosh x.$

Hence the integral is given by

$\int {\sinh x\cosh \left( { - x} \right)dx} = \int {\sinh x\cosh xdx} = \int {\frac{{\sinh 2x}}{2}dx} = \frac{1}{2}\int {\sinh 2xdx} = \frac{1}{2} \cdot \frac{{\cosh 2x}}{2} + C = \frac{{\cosh 2x}}{4} + C.$

### Example 16.

Calculate the integral $\int {\sinh 2x\cosh 3xdx}.$

Solution.

Applying the formulas

$\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} \;\;\text{and}\;\; \cosh x = \frac{{{e^x} + {e^{ - x}}}}{2},$

we get

$\int {\sinh 2x\cosh 3xdx} = \int {\frac{{{e^{2x}} - {e^{ - 2x}}}}{2} \cdot \frac{{{e^{3x}} + {e^{ - 3x}}}}{2}dx} = \frac{1}{4}\int {\left( {{e^{5x}} - {e^x} + {e^{ - x}} - {e^{ - 5x}}} \right)dx} = \frac{1}{4}\left( {\frac{{{e^{5x}}}}{5} - {e^x} - {e^{ - x}} + \frac{{{e^{ - 5x}}}}{5}} \right) + C = \frac{1}{{10}} \cdot \frac{{{e^{5x}} + {e^{ - 5x}}}}{2} - \frac{1}{2} \cdot \frac{{{e^x} + {e^{ - x}}}}{2} + C = \frac{{\cosh 5x}}{{10}} - \frac{{\cosh x}}{2} + C.$

### Example 17.

Find the integral $\int {\tanh 2xdx}.$

Solution.

Given that $$\tanh x = \frac{{\sinh x}}{{\cosh x}},$$ we can write

$I = \int {\tanh 2xdx} = \int {\frac{{\sinh 2x}}{{\cosh 2x}}dx} .$

Make the substitution

$u = \cosh 2x,\;\; \Rightarrow du = 2\sinh 2xdx,\;\; \Rightarrow \sinh 2xdx = \frac{{du}}{2}.$

This yields:

$I = \int {\frac{{\sinh 2x}}{{\cosh 2x}}dx} = \frac{1}{2}\int {\frac{{du}}{u}} = \frac{1}{2}\ln \left| u \right| + C = \ln \sqrt {\left| u \right|} + C = \ln \sqrt {\left| {\cosh 2x} \right|} + C.$

Note that the hyperbolic cosine function is always positive. Hence,

$I = \ln \sqrt {\cosh 2x} + C.$

### Example 18.

Find the integral $\int {\coth \frac{x}{3} dx}.$

Solution.

By definition,

$\coth x = \frac{{\cosh x}}{{\sinh x}}.$

Hence, we can write

$I = \int {\coth \frac{x}{3}dx} = \int {\frac{{\cosh \frac{x}{3}}}{{\sinh \frac{x}{3}}}dx} .$

Use the substitution

$u = \sinh \frac{x}{3},\;\; \Rightarrow du = \frac{1}{3}\cosh \frac{x}{3}dx,\;\; \Rightarrow \cosh \frac{x}{3}dx = 3du.$

Then the integral is given by

$I = \int {\frac{{\cosh \frac{x}{3}}}{{\sinh \frac{x}{3}}}dx} = \int {\frac{{3du}}{u}} = 3\ln \left| u \right| + C = 3\ln \left| {\sinh \frac{x}{3}} \right| + C.$

### Example 19.

Evaluate the integral $\int {\sinh x\cos xdx}.$

Solution.

We use integration by parts. Let

$u = \cos x,\;\;dv = \sinh xdx,\;\; \Rightarrow du = - \sin xdx,\;\; v = \int {\sinh xdx} = \cosh x.$

Then

$\int {\sinh x\cos xdx} = \cosh x\cos x - \int {\cosh x\left( { - \sin x} \right)dx} = \cosh x\cos x + \int {\cosh x\sin xdx}.$

Apply integration by parts again to the latter integral. Let

$u = \sin x,\;\; dv = \cosh xdx,\;\; \Rightarrow du = \cos xdx,\;\; v = \int {\cosh xdx} = \sinh x.$

Then we have

$\int {\sinh x\cos xdx} = \cosh x\cos x + \int {\cosh x\sin xdx} = \cosh x\cos x + \left( {\sin x\sinh x - \int {\sinh x\cos xdx} } \right).$

Solving this equation for $$\int {\sinh x\cos xdx},$$ we obtain the complete answer:

$\int {\sinh x\cos xdx} = \frac{{\cosh x\cos x + \sinh x \sin x}}{2}.$